 Now we can solve some projectile motion problems with the information we know. Again, projectile motion is motion for which the acceleration is gravity, which means it's 9.8 meters per second downward, which gives me my two components of negative 9.8 for the y component and zero for the x component. It's like free fall, but motion in two dimensions. So that means I've got constant acceleration equations for the y direction. Where I can fill in my acceleration of minus 9.8. And I've got zero acceleration for the x direction. So that simplifies my equations just a little bit. For my strategy, I can use the same list that I was using when I was doing one-dimensional constant acceleration. I've got an x and a y components. And like free fall, I can immediately fill in my accelerations with the two values that I know for every projectile type problem. You could also use a simplified x side where you list only the relevant variables, the displacement, the velocity and x in the time. And we don't have to say initial and final velocity because it never changes. And we don't have to list the acceleration because it's not going to come into our equations. In that case, I would only use just the one equation that relates the displacement, the velocity and the time. But I still have a full list of five equations to use for the y side. So now we get to our example. In this example, we've got a water balloon that's being launched up from the ground and we know its speed and the angle that it's being launched at. And we want to know what height the water balloon hits a wall. Well, if I think about my knowns, that speed is the magnitude of the initial velocity. And my angle is known. And that 30 meters is the delta x. Because when we measure how far a wall is, we're measuring the horizontal distance. And the height that I'm measuring, that's the delta y. That's what I'm trying to solve for. Now we need to use trig on that initial velocity because we need the components. So you'll work this out using your normal trig relationships and you can double check these calculations for yourself. So we take our tables and we now fill in the known data. And I'm going ahead and using the full tables just so we can see all the information as it fills in there. You'll notice we have enough information on the x side to start solving over here because I've got my three values. So for my three values over here, I can use my equation for the horizontal displacement. Solve that for the time and fill in my known values. And I end up finding my time of 1.958 seconds. I like to keep a couple extra decimal points around as I'm solving my intermediate steps just so I don't get rounding errors. We can take that time now and transfer it to the y side. So if I got my list for quantities for the y side, now I've got three things and I can solve for my delta y. So I use my vertical displacement equation, fill in all the things I know now and end up finding that I've got 6.39 meters for the height above the ground that the water balloon hit. So that solves my original problem. But I could also ask what's the final speed that that water balloon hits the wall? So you'll notice with all my information filled in now, it's the final velocity components in x and y that I still have to find. Well, if I start with my horizontal final velocity, I find that it's exactly equal to the initial velocity because there is no acceleration. So that component is still 15.32 meters per second. For the vertical velocity, I have an acceleration, so I have to use my full equation here, plug in my initial velocity in the y direction and my acceleration and my time, and I find a negative 6.33 meters per second for the final velocity in the y direction. That means I'm going 6.33 meters per second downward, so the water balloon traveled up to its maximum height and was already on the way down when it hit the wall. To find the overall final speed, again, I need the magnitude of the final velocity. So if I plug those into my typical magnitude formula, squaring each of the components, being careful with my parentheses on the negative sign, I find that I had 16.6 meters per second for my final speed that it hit the wall. So that gives you an example of solving just one example of a projectile motion problem. You'll see lots of different ones where we're given different pieces of knowns and we're solving for different pieces of information.